If $x \boxdot y = 6x-y$ and $x \triangleright y = xy+3x-y$, find $1 \boxdot (6 \triangleright 1)$.
Solution: First, find $6 \triangleright 1$ $ 6 \triangleright 1 = (6)(1)+(3)(6)-1$ $ \hphantom{6 \triangleright 1} = 23$ Now, find $1 \boxdot 23$ $ 1 \boxdot 23 = (6)(1)-23$ $ \hphantom{1 \boxdot 23} = -17$.